A ball of mass M is thrown vertically upwards. Another ball of mass 2M is thrown at an angle `theta` with the vertical. Both of them stay in air for the same period of time. The heights attained by the two are in the ratio

Options

A. `1:2`
B. `2:1`
C. `1:1`
D. `1 : cos theta`

Answer

C

Explanation

From figure Average velocity = displacement time taken by the ball to come back is `T_(1)=(2u_1)/g`
For the ball projected at an angle `theta ` with the vertical , the time of flight is `T_(2) = (2u_(2) cos theta)/g`
Since time for both balls is same , so
`(2u_1)/g=(2u_2cos theta)/g" or " u_1=u_2cos theta " "...(i)`
Now , `h_1=(u_1^2)/(2g) and h_2 =(u_2^2)/(2g)cos^(2) theta `
Hence , `h_1/h_2=(u_1^2)/(u_2^2cos^2theta)=(u_2^2cos^(2) theta)/(u_2^2cos^(2) theta)=1 " "` (Using (i))